Timeline for equivariant cohomology
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2012 at 15:06 | history | edited | user15512 | CC BY-SA 3.0 |
LaTeX correction; added 8 characters in body; added 8 characters in body; added 11 characters in body; deleted 11 characters in body
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| Jul 10, 2012 at 11:48 | comment | added | Sean Tilson | Sometimes Borel cohomology is not the right equivariant theory though. I was assuming he was looking at some $RO(G)$ graded theory. Maybe that could be clarified in the original question. Maybe it is obvious to the relevant parties though. | |
| Jul 9, 2012 at 22:38 | comment | added | Chris Gerig | @Ben, this is standard notation for equivariant cohomology, $H_G^\ast(M)=H^\ast(M_G)$ where $M_G$ is the Borel construction. | |
| Jul 9, 2012 at 22:15 | comment | added | DamienC | What does this mean when $G/H$ is not a group ? | |
| Jul 9, 2012 at 22:04 | answer | added | Ralph | timeline score: 6 | |
| Jul 9, 2012 at 21:59 | comment | added | Ben Webster♦ | Could you explain the notation $H^*_{G/H}(M)$? | |
| Jul 9, 2012 at 21:53 | history | edited | MTS | CC BY-SA 3.0 |
Fixed latex, removed group-cohomology tag
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| Jul 9, 2012 at 21:38 | comment | added | user15512 | Assuming that the cohomology is with rational coefficients, I am hoping for some relation that would determine the ring $H^*_{G/H}(M)$ in terms of $H^*_G (M)$ and $H^*_{G/H}$. For instance, when is $H^*_{G/H}(M)$ isomorphic as rings to $H^*_{G}(M)\otimes H^*_{G/H}$ ? | |
| Jul 9, 2012 at 21:25 | comment | added | Yemon Choi | What kind of relationship do you want? (I am not being flippant; it sometimes - often? - pays in mathematics to have some idea of what one hopes to be true, before one tries to invent or look up a proof.) | |
| Jul 9, 2012 at 21:22 | history | asked | user15512 | CC BY-SA 3.0 |